Download Orders in Self-lnjective Semi-Perfect Rings

JOURNAL
OF ALGEBRA
13,
Orders
of North
(1969)
in Self-lnjective
A.
University
5-g
C. MEWBORN
AND
Department
Carolina
at Chapel
Communicated
Received
Semi-Perfect
C. N.
Rings
WINTON
of Mathematics,
Hill,
Chapel
Hill,
by Nathalt
Jacobsox
March
1968
15,
North
Carolina
27514
1. INTRODUCTION
If R is a subring of a ring Q with unity (and R contains
then R is a right order in Q if
a regular
(a) every regular element of R is a unit in Q, and
(b) every element 4 EQ can be written 4 = ab-r where
is regular in R.
element)
a, b E R and b
In [2] Jans proved that the following conditions are necessary and suflicient
for a ring R containing
a regular element to be a right order in a quasiFrobenius
ring:
(1) R, is a finite dimensional
sum of nonzero right ideals.
module;
i.e., R contains
no infinite
direct
(2) If x EI(RJ, the injective hull of RR , then there is a regular element
b E R such that xb ER.
(3) R satisfies the ascending chain condition on annihilators
of subsets of
4&J(4) If M is a finitely generated (or cyclic) R-module
such that mb f 0
for each TPZE M if b is regular in R, then the injective hull of M is contained
in the product of a finite number of copies of I(R,j.
In this paper we sharpen the result of Jans by showing that conditions
(l), (2) and (3) are sufficient for R to be a right order in a quasi-Frobenius
ring. Conditions
(1) and (2) are necessary and sufficient that R be a right
order in a self-injective
semi-perfect
ring. Let (3’) be the condition:
(3’)
R has ascending
chain condition
on annihilator
right
ideals.
If R satisfies (l), (2), and (3’) then R is a right order in a self-injective
semi-primary
ring. Conditions
(l), (2), (3), (3’), and (4) will be referred to by
number in this paper.
5
6
MEWBORN
Assume
a regular
TR(M) =
following
AND
WINTON
that R is a ring (not necessarily with unity) and that R contains
element. Following
Jans [2], if M is a (right) R-module
then
{m E M : mb = 0, some regular element 6 E R}. We need the
lemma:
LEMMA
2.1. Let M be an R-module such that TR(M) = (0) and assume
that for every right ideal H of R each.R-map H -+ M can be extendedto an
R-map R + M. Then M is inj*ective.(The reader will note this is1.usta modification of Baer’s criterion for a moduleover a ring with unity to be injective.)
Proof. Let A, B be R-modules,
A C B, and let 4 : A --f M. If 4
cannot be extended to B, by use of Zorn’s lemma we may assume 4 cannot
be extended to any R-module
C, A C C 2 B. Let b EB, b $ A.
H={rERIbrEA}
is then a right ideal and we can define f : H -+ M by ,f(r) = $(br). Hence,
f can be extended to f : R -+ M by hypothesis. If H C R, the R-module
by
C={a+brIaEA,rER}r)A
and we can defme $:C+M
since a + br = 0 * br E A 3
$<a + bf-1 = d(a) +Jk>(4 IS
* well-defined
f(r) =f(r) = $(br) so that $(a + br) = $(a + by) = 0). If H = R,
C = {a + nb 1 a E A, n E Z> is an R-module
properly containing
A. Let d
be regular in R. Then the map # : R -+ dR given by C)(Y) = dr is manic and
so there exists 4-l : dR -+ R. Then th e composition f 0 z,P can be extended
to 0:R-+M.
Define
4 : C + M by $(a + nb) = $(a) + m?(d). If
a + nb = 0, nb E A and so $(&I) d = +(nbd) = f(nd) = (nB(d2)) = ne(d)d.
Hence, $(nb) = m?(d)since TR(M) = 0. Then&a
+ nb) = $(a + nb) = 0;
i.e., 4 is well-defined.
Hence, 4 can be extended to B; i.e., M is injective.
LEMMA
2.2. Let R be a r&ht order in a selfinjective ring Q. Then Q is
R-injective.
Proof. Let R be the ring with unity generated by R and the unity of Q.
Then Q is a self-injective
ring of right quotients of R and so must be the
complete ring of quotients of R. (See [3]: $4.3). Hence QE is the injective
hull of R. Any R-map from H to Q is an R-map and so extends to an R-map
R + Q; hence, an R-map. Clearly TR(Q) = (0). Thus QR is injective by
Lemma 2.1. The Lemma is proved.
If R is a right order in Q, Qs is an essential extension of RR , hence
QR =
ERR)-
Since
HOmR(RR
R contains a regular element we can embed R naturally
into
, RR) by mapping r E R to the left multiplication
by Y. Let R
ORDERS
IN
SELF-INJECTIVE
SEMI-PERFECT
RINGS
7
be the subring of HomR(R,
, RR) generated by the image of R and the
identity map. RR is large in Hom,(R,
, RR) and hence in RR. Let I = f&)
denote the injective hull of i?~ ( as an R-module).
It is easily seen that I is an
essential extension of RR (as an R-module).
PROPOSITION
2.3. If R satisfies conditions (1) and Q) of the introduction,
i.e., RR is finite dimensional and TR(I(RR)/RR)
= I(RR)/RR , then the complete
ring of quotients Q of i? is self-injective. Q can be identijied with I(&)
as an
R-module and with I(R,) as an R-module.
Proof.
We show first that if h E HomR(1,1)
and h(a) = (0) then h = 0.
Suppose h E HomK(I, I), h f 0. Th en h(I) is a nonzero R-submodule,
and
hence an R-submodule,
of I. Thus h(I) n R f (0). Let i E.Z such that
h(i) + 0, h(i) E R. By assumption there is a regular element 6 E R such that
ib E R. Then h(ib) = h(i) 6 # 0. Hence h(R) f (0). By ([3] : $4.3) the
complete ring of quotients Q of R is self-injective,
and, QR can be identified
with I(&).
We now show that Q is R-injective.
Let b be a regular element of R, and let & : Q -+ Q be the left multiplication by b; & : x -+ bx, x EQ. Since the intersection
of R with the kernel
of &, is zero and Q is an essential extension of R, &, is a monomorphism.
&(Q) is an R-injective
submodule
of Q and so it is a direct summand:
Q = 4&Q) @ K. If K f (0), we get for each positive integer n,
Q = #bYQ) 0 K1(K)
0 *-* 0 K.
This implies that Q is not finite dimensional
as R-module,
hence also as
R-module.
QR is an essential extension of R, , hence RR is not finite
dimensional,
contradicting
Condition
(1). Therefore
K = (0) and & is an
R-isomorphism
of Q onto Q; i.e., b is invertible in Q.
Since each regular element of R is invertibIe in Q it is clear that TR(Q) = (0).
If H is a right ideal of R and f : H + Q is an R-map, then f is also an R-map.
Hence f extends to an R-map f’ : Q --j Q. f’ is also an R-map. By Lemma 2.1,
QR is R-injective,
and hence may be identified with I(R,). This completes the
proof of the proposition.
THEOREM
2.4. R is a right oTdel- in a selfinjective
if and only if R satis$es conditions (1) and (2).
semi-perfect
ring Q
Proof.
Assume R is a right order in a self-injective
semi-perfect
ring Q.
Then Qa is the injective hull of RR . Q is the direct sum of finitely many
indecomposable
Q-injective
right ideals, Q = Qr @ =.a @ QTz . Each direct
summand Qi is R-injective
and is in fact the injective hull of Q$ n R which
is a uniform right ideal of R. Hence R is finite dimensional.
If q E Q then
q = ab-I, a, b E R, b regular. Hence qb E R. Thus R satisfies Condition
(2).
8
iVI!JWBORN
AND
WINTON
Conversely,
suppose R satisfies Conditions
(1) and (2). The complete
ring of quotients Q of R is self injective. Since R is finite dimensional
so is i?.
By ([3] : p. 103) Q is semi-perfect.
Let b E R, b regular. As in the proof of
Proposition
2.3. we see that b is invertible in Q. If ~1E Q then by Condition
(2) there exists a regular element b E R such that $ = a E R. Then 4 = ab-r.
Therefore R is a right order in Q.
THEOREM2.5. If R satisjiesConditions(l), (2), and (3’) then R is a @&t
order in a seEf&jective semi-primaryring Q.
Proof. By Theorem 2.4 it is necessary only to show that Rad Q is nilpotent.
By ([3] : p. 102) Rad Q is the (right) singular ideal of Q and is the singular
R-submodule
of Qx . We show first that the singular ideal J of R is nilpotent.
Suppose I”+’ f (0), k a positive integer. Among all elements a E J such that
]“a f (0) choose a such that its right annihilator
(a)Y is maximal. If 6 E J
then (6)v n aR+ (0). Hence there exists Y E R such that ar + 0 but bar = 0.
Thus (a)YC (bay. H ence Jkba = (0), all 6 E J; i.e. ]“+‘a = (0). If Jwere not
nilpotent we would have
an infinite ascending chain of annihilator
right ideals, which is not possible.
Thus J is nilpotent.
Now
J = Rad Q n R. Suppose
J” = (0). Let qi = a$;’ E Rad Q,
a; , bi E R, 6, regular, 1 < i < n. Then qrqa a** qn can be written in the form
al’a2’ *ema,‘b-l where ai’ E J, b E R, b regular. Hence %‘a,’ me*a,‘b-l = 0,
and so qlqz **a qa = 0. It follows that (RadQ)”
= (0).
THEOREM
2.6. R is a rigrt order in a quasi-Frobenius
ring Q if and only if R
satisfiesConditions(I), (2), and (3).
Proof. If R is a right order in a quasi-Frobenius
ring then R satisfies
Conditions (l), (2), and (3) by [2]. Supp ose R satisfies Conditions (l), (2), and
(3). By Theorem 2.4 R is a right order in a self-injective
semi-perfect ring Q.
By ([I] : 5.2) Q is quasi-Frobenius
if it satisfies the ascending chain condition
on annihilator
right ideals. Suppose X, , X, CQ, and
{qEQ: X,q = (O)}C(qEQ:
Let q = ab-l be such that Xgq = (0), X,q f
Then xla f 0 but X,a = (0). Hence
X,q = (0)).
(0). Let x1 E X1 , xlq # 0.
(r E R : X,y = (0)) C {r ER : X,r = (0)).
Therefore
condition
it follows from Condition
(3) that Q satisfies the ascending
on right ideals. The theorem is proved.
chain
ORDERS
Note
Osofsky
IN
SELF-INJECTIVE
added in proof.
The authors
have
have also (independently)
observed
SEMI-PERFECT
learned
that
that condition
9
RINGS
Lance Small
and Barbara
(4) of Jans is superfluous.
1. FAITH, C. Rings with ascending
condition
on annihilators.
Nagoya
(1966),
179-191.
2. JANS, J. P. On orders in quasi-frobenius
rings. J. Algebra
7 (1967),
3. LAMBEK,
J. “Lectures
on Rings and Modules.”
Ginn
(Blaisdell),
Math.
J. 27
35-43.
Boston,
1966.